Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		double p[] = new double[n + 1];
		double q[] = new double[n + 1];
		for ( int i = 1; i <= n; i++ ) {
			p[i] = in.nextDouble();
			q[i] = 1 - p[i];
		}
		in.close();

		//		for(int i = 0; i<= n; i++) {
		//			System.out.println(p[i] + "  " + q[i]);
		//		}

		int row = n / 2 + 1;
		double dp[][] = new double[row + 1][n + 2];
		dp[1][1] = 1;
		for ( int r = 1; r <= row; r++ ) {
			for ( int c = 1; c <= n + 1; c++ ) {
				int idx = r + c - 2;
				dp[r][c] += dp[r][c - 1] * p[idx];
				dp[r][c] += dp[r - 1][c] * q[idx];
				if ( idx >= n ) break;
			}
		}

//		for ( double[] ds : dp ) {
//			for ( double d : ds ) {
//				System.out.printf("%1.3f   ", d);
//			}
//			System.out.println();
//		}

		double ans = 0.0;
		for ( int r = 1, c = n + 1; r <= row; r++, c-- ) {
			ans += dp[r][c];
		}
		System.out.println(ans);
	}
}